Respuesta :
Explanation:
Expression for energy balance is as follows.
[tex]\Delta E_{system} = E_{in} - E_{out}[/tex]
or, [tex]E_{in} = E_{out}[/tex]
Therefore,
[tex]m(h_{1} \frac{v^{2}_{1}}{2}) = m (h_{2} \frac{V^{2}_{2}}{2})[/tex]
[tex]h_{1} + \frac{V^{2}_{1}}{2} = h_{2} + \frac{V^{2}_{2}}{2}[/tex]
Hence, expression for exit velocity will be as follows.
[tex]V_{2} = [V^{2}_{1} + 2(h_{1} - h_{2})^{0.5}[/tex]
= [tex]V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}[/tex]
As [tex]C_{p}[/tex] for the given conditions is 1.007 kJ/kg K. Now, putting the given values into the above formula as follows.
[tex]V_{2} = V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}[/tex]
= [tex][(350 m/s)^{2} + 2(1.007 kJ/kg K) (30 - 90) K \frac{1000 m^{2}/s^{2}}{1 kJ/kg}]^{0.5}[/tex]
= 40.7 m/s
Thus, we can conclude that velocity at the exit of a diffuser under given conditions is 40.7 m/s.
